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# Oscillatory behaviour of a higher-order dynamic equation

Journal of Inequalities and Applications20132013:52

https://doi.org/10.1186/1029-242X-2013-52

• Received: 26 September 2012
• Accepted: 30 January 2013
• Published:

## Abstract

In this paper we are concerned with the oscillation of solutions of a certain more general higher-order nonlinear neutral-type functional dynamic equation with oscillating coefficients. We obtain some sufficient criteria for oscillatory behaviour of its solutions.

MSC:34N05.

## Keywords

• time scale
• higher-order nonlinear neutral dynamic equation
• oscillating coefficient

## 1 Introduction

The calculus on time scales has been introduced in order to unify the theories of continuous and discrete processes and in order to extend those theories to a more general class of the so-called dynamic equations. In recent years there has been much research activity concerning the oscillation and non-oscillation of solutions of neutral dynamic equations on time scales.

In this paper we consider the higher-order nonlinear dynamic equation
${\left[y\left(t\right)+P\left(t\right)y\left(\tau \left(t\right)\right)\right]}^{{\mathrm{\Delta }}^{n}}+\sum _{i=1}^{m}{Q}_{i}\left(t\right){f}_{i}\left(y\left({\varphi }_{i}\left(t\right)\right)\right)=0,$
(1.1)

where $n\ge 2$, $P\left(t\right),{Q}_{i}\left(t\right)\in {C}_{rd}{\left[{t}_{0},\mathrm{\infty }\right)}_{\mathbb{T}}$ for $i=1,2,\dots ,m$; $P\left(t\right)$ is an oscillating function ($P\left(t\right):\mathbb{T}\to \mathbb{R}$), ${Q}_{i}\left(t\right)$ are positive real-valued functions for $i=1,2,\dots ,m$; ${\varphi }_{i}\left(t\right)\in {C}_{rd}{\left[{t}_{0},\mathrm{\infty }\right)}_{\mathbb{T}}$, ${\varphi }_{i}^{\mathrm{\Delta }}\left(t\right)>0$, the variable delays $\tau ,{\varphi }_{i}:{\left[{t}_{0},\mathrm{\infty }\right)}_{\mathbb{T}}\to \mathbb{T}$ with $\tau \left(t\right),{\varphi }_{i}\left(t\right) for all $t\in {\left[{t}_{0},\mathrm{\infty }\right)}_{\mathbb{T}}$, ${\varphi }_{i}\left(t\right)\to \mathrm{\infty }$ as $t\to \mathrm{\infty }$ for $i=1,2,\dots ,m$; $\tau \left(t\right)\to \mathrm{\infty }$ as $t\to \mathrm{\infty }$; ${f}_{i}\left(u\right)\in C\left(\mathbb{R},\mathbb{R}\right)$ are nondecreasing functions, $u{f}_{i}\left(u\right)>0$ for $u\ne 0$ and $i=1,2,\dots ,m$.

The purpose of the paper is to study oscillatory behaviour of solutions of equation (1.1). For the sake of convenience, the function $z\left(t\right)$ is defined by
$z\left(t\right)=y\left(t\right)+P\left(t\right)y\left(\tau \left(t\right)\right).$
(1.2)

## 2 Basic definitions and some auxiliary lemmas

A time scale $\mathbb{T}$ is an arbitrary nonempty closed subset of the real numbers . For $t\in \mathbb{T}$, we define the forward jump operator $\sigma :\mathbb{T}\to \mathbb{T}$ by
$\sigma \left(t\right):=inf\left\{s\in \mathbb{T}:s>t\right\}$
while the backward jump operator $\rho :\mathbb{T}\to \mathbb{T}$ is defined by
$\rho \left(t\right):=sup\left\{s\in \mathbb{T}:s
If $\sigma \left(t\right)>t$, we say that t is right-scattered, while if $\rho \left(t\right), we say that t is left-scattered. Also, if $\sigma \left(t\right)=t$, then t is called right-dense, and if $\rho \left(t\right)=t$, then t is called left-dense. The graininess function $\mu :\mathbb{T}\to \left[0,\mathrm{\infty }\right)$ is defined by
$\mu \left(t\right):=\sigma \left(t\right)-t.$

We introduce the set ${\mathbb{T}}^{\kappa }$ which is derived from the time scale $\mathbb{T}$ as follows. If $\mathbb{T}$ has left-scattered maximum m, then ${\mathbb{T}}^{\kappa }=\mathbb{T}-\left\{m\right\}$, otherwise ${\mathbb{T}}^{\kappa }=\mathbb{T}$.

Definition 1 

The function $f:\mathbb{T}\to \mathbb{R}$ is called rd-continuous provided it is continuous at right-dense points in $\mathbb{T}$ and its left-sided limits exist (finite) at left-dense points in $\mathbb{T}$.

Theorem 1 

Assume that $\nu :\mathbb{T}\to \mathbb{R}$ is strictly increasing and $\stackrel{˜}{\mathbb{T}}:=\nu \left(\mathbb{T}\right)$ is a time scale. Let $w:\stackrel{˜}{\mathbb{T}}\to \mathbb{R}$. If ${\nu }^{\mathrm{\Delta }}\left(t\right)$ and ${w}^{\stackrel{˜}{\mathrm{\Delta }}}\left(\nu \left(t\right)\right)$ exist for $t\in {\mathbb{T}}^{\kappa }$, then
${\left(w\circ \nu \right)}^{\mathrm{\Delta }}=\left({w}^{\stackrel{˜}{\mathrm{\Delta }}}\circ \nu \right){\nu }^{\mathrm{\Delta }},$

where we denote the derivative on $\stackrel{˜}{\mathbb{T}}$ by $\stackrel{˜}{\mathrm{\Delta }}$.

Definition 2 

Let $f:\mathbb{T}\to \mathbb{R}$ be a function. If there exists a function $F:\mathbb{T}\to \mathbb{R}$ such that ${F}^{\mathrm{\Delta }}\left(t\right)=f\left(t\right)$ for all $t\in {\mathbb{T}}^{\kappa }$, then F is said to be an antiderivative of f. We define the Cauchy integral by

Theorem 2 

Let u and v be continuous functions on $\left[a,b\right]$ that are Δ-differentiable on $\left[a,b\right)$. If ${u}^{\mathrm{\Delta }}$ and ${v}^{\mathrm{\Delta }}$ are integrable from a to b, then
${\int }_{a}^{b}{u}^{\mathrm{\Delta }}\left(t\right)v\left(t\right)\mathrm{\Delta }\left(t\right)+{\int }_{a}^{b}{u}^{\sigma }\left(t\right){v}^{\mathrm{\Delta }}\left(t\right)\mathrm{\Delta }\left(t\right)=u\left(b\right)v\left(b\right)-u\left(a\right)v\left(a\right).$
Let $\stackrel{˜}{\mathbb{T}}=\mathbb{T}\cup \left\{sup\mathbb{T}\right\}\cup \left\{inf\mathbb{T}\right\}$. If $\mathrm{\infty }\in \stackrel{˜}{\mathbb{T}}$, we call ∞ left-dense, and −∞ is called right-dense provided $-\mathrm{\infty }\in \stackrel{˜}{\mathbb{T}}$. For any left-dense ${t}_{0}\in \stackrel{˜}{\mathbb{T}}$ and any $\epsilon >0$, the set
${L}_{\epsilon }\left({t}_{0}\right)=\left\{t\in \mathbb{T}:0<{t}_{0}-t<\epsilon \right\}$

is nonempty, and so is ${L}_{\epsilon }\left(\mathrm{\infty }\right)=\left\{t\in \mathbb{T}:t>\frac{1}{\epsilon }\right\}$ if $\mathrm{\infty }\in \stackrel{˜}{\mathbb{T}}$.

Lemma 1 

Let $n\in \mathbb{N}$ and f be n-times differentiable on $\mathbb{T}$. Assume $\mathrm{\infty }\in \stackrel{˜}{\mathbb{T}}$. Suppose there exists $\epsilon >0$ such that
$f\left(t\right)>0,\phantom{\rule{1em}{0ex}}sgn\left({f}^{{\mathrm{\Delta }}^{n}}\left(t\right)\right)\equiv s\in \left\{-1,+1\right\}\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.1em}{0ex}}t\in {L}_{\epsilon }\left(\mathrm{\infty }\right)$
and ${f}^{{\mathrm{\Delta }}^{n}}\left(t\right)\ne 0$ on ${L}_{\delta }\left(\mathrm{\infty }\right)$ for any $\delta >0$. Then there exists $p\in \left[0,n\right]\cap {\mathbb{N}}_{0}$ such that $n+p$ is even for $s=1$ and odd for $s=-1$ with
$\left\{\begin{array}{l}{\left(-1\right)}^{p+j}{f}^{{\mathrm{\Delta }}^{j}}\left(t\right)>0\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.1em}{0ex}}t\in {L}_{\epsilon }\left(\mathrm{\infty }\right),j\in \left[p,n-1\right]\cap {\mathbb{N}}_{0},\\ {f}^{{\mathrm{\Delta }}^{j}}\left(t\right)>0\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.1em}{0ex}}t\in {L}_{{\delta }_{j}}\left(\mathrm{\infty }\right)\phantom{\rule{0.25em}{0ex}}\left(\mathit{\text{with}}\phantom{\rule{0.1em}{0ex}}{\delta }_{j}\in \left(0,\epsilon \right)\right),j\in \left[1,p-1\right]\cap {\mathbb{N}}_{0}.\end{array}$

Lemma 2 

Let f be n-times differentiable on ${\mathbb{T}}^{{\kappa }^{n}}$, $t\in \mathbb{T}$, and $\alpha \in {\mathbb{T}}^{{\kappa }^{n}}$. Then with the functions ${h}_{k}$ defined as ${h}_{n}\left(t,s\right)={\left(-1\right)}^{n}{g}_{n}\left(s,t\right)$,
${h}_{0}\left(r,s\right)\equiv 1\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{h}_{k+1}\left(r,s\right)={\int }_{s}^{r}{h}_{k}\left(\tau ,s\right)\mathrm{\Delta }s\phantom{\rule{1em}{0ex}}\mathit{\text{for}}\phantom{\rule{0.1em}{0ex}}k\in {\mathbb{N}}_{0},$
we have
$f\left(t\right)=\sum _{k=0}^{n-1}{h}_{k}\left(t,\alpha \right){f}^{{\mathrm{\Delta }}^{k}}\left(\alpha \right)+{\int }_{\alpha }^{{\rho }^{n-1}\left(t\right)}{h}_{n-1}\left(t,\sigma \left(\tau \right)\right){f}^{{\mathrm{\Delta }}^{n}}\left(\tau \right)\mathrm{\Delta }\tau .$

Lemma 3 

Let f be n-times differentiable on ${\mathbb{T}}^{{\kappa }^{n}}$ and $m\in \mathbb{N}$ with $m. Then we have, for all $\alpha \in {\mathbb{T}}^{{\kappa }^{n-1+m}}$ and $t\in {\mathbb{T}}^{{\kappa }^{m}}$,
${f}^{{\mathrm{\Delta }}^{m}}\left(t\right)=\sum _{k=0}^{n-m-1}{h}_{k}\left(t,\alpha \right){f}^{{\mathrm{\Delta }}^{k+m}}\left(\alpha \right)+{\int }_{\alpha }^{{\rho }^{n-m-1}\left(t\right)}{h}_{n-m-1}\left(t,\sigma \left(\tau \right)\right){f}^{{\mathrm{\Delta }}^{n}}\left(\tau \right)\mathrm{\Delta }\tau .$

Lemma 4 

Suppose f is n-times differentiable and ${g}_{k}$, $0\le k\le n-1$, are differentiable at $t\in {\mathbb{T}}^{{\kappa }^{n}}$ with
${g}_{k+1}^{\mathrm{\Delta }}\left(t\right)={g}_{k}\left(\sigma \left(t\right)\right)\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.1em}{0ex}}0\le k\le n-2.$
Then we have
${\left[\sum _{k=0}^{n-1}{\left(-1\right)}^{k}{f}^{{\mathrm{\Delta }}^{k}}{g}_{k}\right]}^{\mathrm{\Delta }}=f{g}_{0}^{\mathrm{\Delta }}+{\left(-1\right)}^{n-1}{f}^{{\mathrm{\Delta }}^{n}}{g}_{n-1}^{\sigma }.$

## 3 Main results

Lemma 5 Let f be n-times differentiable on ${\mathbb{T}}^{{\kappa }^{n}}$. If ${f}^{\mathrm{\Delta }}>0$, then for every λ, $0<\lambda <1$, we have
$f\left(t\right)\ge \lambda {\left(-1\right)}^{n-1}{g}_{n-1}\left(\sigma \left({T}^{\ast }\right),t\right){f}^{{\mathrm{\Delta }}^{n-1}}\left(t\right).$
(3.1)
Proof Let p, $0\le p\le n-1$, be the integer assigned to the function f as in Lemma 1. Because of ${f}^{\mathrm{\Delta }}>0$, we always have $p>0$. Furthermore, let ${T}^{\ast }\ge T$ be assigned to f by Lemma 1. Then, by using the Taylor formula on time scales, for every ${\rho }^{n-1}\left(t\right)\ge {T}^{\ast }$, we obtain
$f\left(t\right)\ge {\int }_{{T}^{\ast }}^{{\rho }^{n-1}\left(t\right)}{\left(-1\right)}^{n-1}{g}_{n-1}\left(\sigma \left(\tau \right),t\right){f}^{{\mathrm{\Delta }}^{n}}\left(\tau \right)\mathrm{\Delta }\tau .$
(3.2)
By using Theorem 2 and (3.2), we have
$f\left(t\right)\ge {\left(-1\right)}^{n-1}{g}_{n-1}\left(\sigma \left(t\right),t\right){f}^{{\mathrm{\Delta }}^{n-1}}\left(t\right)-{\int }_{{T}^{\ast }}^{{\rho }^{n-1}\left(t\right)}{\left(-1\right)}^{n-1}{g}_{n-1}\left(\sigma \left(\tau \right),t\right){f}^{{\mathrm{\Delta }}^{n-1}}\left(\tau \right)\mathrm{\Delta }\tau .$
Since f is n-times differentiable on ${\mathbb{T}}^{{\kappa }^{n}}$ and $m\in \mathbb{N}$ with $m, we have with n and f substituted by $n-m$ and ${f}^{{\mathrm{\Delta }}^{m}}$, respectively
${f}^{{\mathrm{\Delta }}^{m}}\left(t\right)\ge {\int }_{{T}^{\ast }}^{{\rho }^{n-m-1}\left(t\right)}{\left(-1\right)}^{n-m-1}{g}_{n-m-1}\left(\sigma \left(\tau \right),t\right){f}^{{\mathrm{\Delta }}^{n}}\left(\tau \right)\mathrm{\Delta }\tau .$
Also, for every ${\rho }^{n-1}\left(t\right)$, s with ${\rho }^{n-1}\left(t\right)\ge {T}^{\ast }$ and ${T}^{\ast }\le s\le t$, we have
${f}^{{\mathrm{\Delta }}^{m}}\left(s\right)\ge {\left(-1\right)}^{n-m-1}{g}_{n-m-1}\left(\sigma \left({T}^{\ast }\right),t\right){f}^{{\mathrm{\Delta }}^{n}}\left(t\right).$
This is obvious for $m=n-1$ and, when $m, it can be derived by applying the Taylor formula. Thus, for all $t\ge {T}^{\ast }$, we get
$f\left(t\right)\ge {\left(-1\right)}^{n-1}{g}_{n-1}\left(\sigma \left({T}^{\ast }\right),t\right){f}^{{\mathrm{\Delta }}^{n-1}}\left(t\right)$

and therefore the proof of the lemma can be immediately completed. □

The result of Lemma 5 is an extension of studies in  and . In order that the reader sees how the results in  (1.8.14) and  (Lemma 2) follow from (3.1), it is at this point only necessary to know that in the case $\mathbb{T}=\mathbb{Z}$, we have $\rho \left(t\right)=t-1$, $\sigma \left(t\right)=t+1$ and
${g}_{n-1}\left(\sigma \left({T}^{\ast }\right),t\right)=\frac{{\left(t-{T}^{\ast }-1\right)}^{\left(n-1\right)}}{\left(n-1\right)!},$
then we get the inequality in 
$u\left(t\right)\ge \frac{1}{\left(n-1\right)!}{\left(n-{n}_{1}\right)}^{\left(n-1\right)}{\mathrm{\Delta }}^{n-1}u\left({2}^{n-m-1}n\right);$
and in the case $\mathbb{T}=\mathbb{R}$, we have $\rho \left(t\right)=\sigma \left(t\right)=t$ and
${g}_{n-1}\left(\sigma \left({T}^{\ast }\right),t\right)=\frac{{\left(t-{T}^{\ast }\right)}^{\left(n-1\right)}}{\left(n-1\right)!},$
then we get the inequality in 
$u\left(t\right)\ge \frac{\vartheta }{\left(n-1\right)!}{\left(t\right)}^{n-1}{u}^{n-1}\left(t\right).$

For the cases $\mathbb{T}=\mathbb{Z}$ and $\mathbb{T}=\mathbb{R}$, some sufficient criterias for oscillatory behaviour of the solutions of the equation (1.1) were obtained by Bolat and Akın in  and , respectively. Furthermore, there might be other time scales that we cannot appreciate at this time due to our current lack of ‘real-world’ examples.

Theorem 3 Assume that n is odd and

(C1) ${lim}_{t\to \mathrm{\infty }}P\left(t\right)=0$,

(C2) ${\int }_{{t}_{0}}^{\mathrm{\infty }}{s}^{n-1}{\sum }_{i=1}^{m}{Q}_{i}\left(s\right)\mathrm{\Delta }s=\mathrm{\infty }$.

Then every bounded solution of equation (1.1) is either oscillatory or tends to zero as $t\to \mathrm{\infty }$.

Proof Assume that equation (1.1) has a bounded non-oscillatory solution $y\left(t\right)$. Without loss of generality, assume that $y\left(t\right)$ is eventually positive (the proof is similar when $y\left(t\right)$ is eventually negative). That is, $y\left(t\right)>0$, $y\left(\tau \left(t\right)\right)>0$ and $y\left({\varphi }_{i}\left(t\right)\right)>0$ for $t\ge {t}_{1}\ge {t}_{0}$ and $i=1,2,\dots ,m$. Assume further that $y\left(t\right)$ does not tend to zero as $t\to \mathrm{\infty }$. By (1.1), (1.2), we have for $t\ge {t}_{1}$
${z}^{{\mathrm{\Delta }}^{n}}\left(t\right)=-\sum _{i=1}^{m}{Q}_{i}\left(t\right){f}_{i}\left(y\left({\varphi }_{i}\left(t\right)\right)\right)<0.$
(3.3)

It follows that ${z}^{{\mathrm{\Delta }}^{j}}\left(t\right)$ ($j\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$) is strictly monotone and eventually of constant sign. Since $P\left(t\right)$ is an oscillatory function, there exists a ${t}_{2}\ge {t}_{1}$ such that if $t\ge {t}_{2}$, then $z\left(t\right)>0$. Since $y\left(t\right)$ is bounded, by virtue of (C1) and (1.2), there is a ${t}_{3}\ge {t}_{2}$ such that $z\left(t\right)$ is also bounded for $t\ge {t}_{3}$. Because n is odd and $z\left(t\right)$ is bounded, by Lemma 1, when $p=0$ (otherwise $z\left(t\right)$ is not bounded), there exists ${t}_{4}\ge {t}_{3}$ such that for $t\ge {t}_{4}$ we have ${\left(-1\right)}^{j}{z}^{{\mathrm{\Delta }}^{j}}\left(t\right)>0$, $j\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$.

In particular, since ${z}^{\mathrm{\Delta }}\left(t\right)<0$ for $t\ge {t}_{4}$, $z\left(t\right)$ is decreasing. Since $z\left(t\right)$ is bounded, we write ${lim}_{t\to \mathrm{\infty }}z\left(t\right)=L$ ($-\mathrm{\infty }). Assume that $0\le L<\mathrm{\infty }$. Let $L>0$. Then there exists a constant $c>0$ and a ${t}_{5}\ge {t}_{4}$ such that $z\left(t\right)>c>0$ for $t\ge {t}_{5}$. Since $y\left(t\right)$ is bounded, ${lim}_{t\to \mathrm{\infty }}P\left(t\right)y\left(\tau \left(t\right)\right)=0$ by (C1). Therefore, there exists a constant ${c}_{1}>0$ and a ${t}_{6}\ge {t}_{5}$ such that $y\left(t\right)=z\left(t\right)-P\left(t\right)y\left(\tau \left(t\right)\right)>{c}_{1}>0$ for $t\ge {t}_{6}$. So that we can find a ${t}_{7}$ with ${t}_{7}\ge {t}_{6}$ such that $y\left({\varphi }_{i}\left(t\right)\right)>{c}_{1}>0$ for $t\ge {t}_{7}$. From (3.3) we have
${z}^{{\mathrm{\Delta }}^{n}}\left(t\right)=-\sum _{i=1}^{m}{Q}_{i}\left(t\right){f}_{i}\left({c}_{1}\right)<0$
(3.4)
for $t\ge {t}_{7}$. If we multiply (3.4) by ${t}^{n-1}$ and integrate it from ${t}_{7}$ to t, we obtain
$F\left(t\right)-F\left({t}_{7}\right)\le -f\left({c}_{1}\right){\int }_{{t}_{7}}^{t}\sum _{i=1}^{m}{Q}_{i}\left(s\right){s}^{n-1}\mathrm{\Delta }s,$
(3.5)
where
$F\left(t\right)=\sum _{i=1}^{n-1}{\left(-1\right)}^{i+1}{\left({t}^{n-1}\right)}^{{\mathrm{\Delta }}^{i}}{z}^{{\mathrm{\Delta }}^{n-i}}\left({\sigma }^{i}\left(t\right)\right)$
and
${\sigma }^{i}\left(t\right)=\sigma \left({\sigma }^{i-1}\left(t\right)\right).$
Since ${\left(-1\right)}^{k}{z}^{{\mathrm{\Delta }}^{k}}\left(t\right)>0$ for $k=0,1,2,\dots ,n-1$ and $t\ge {t}_{4}$, we have $F\left(t\right)>0$ for $t\ge {t}_{7}$. From (3.5) we have
$-F\left({t}_{7}\right)\le -f\left({c}_{1}\right){\int }_{{t}_{7}}^{t}\sum _{i=1}^{m}{Q}_{i}\left(s\right){s}^{n-1}\mathrm{\Delta }s.$
By (C2) we obtain
$-F\left({t}_{7}\right)\le -f\left({c}_{1}\right){\int }_{{t}_{7}}^{t}\sum _{i=1}^{m}{Q}_{i}\left(s\right){s}^{n-1}\mathrm{\Delta }s=-\mathrm{\infty }$
as $t\to \mathrm{\infty }$. This is a contradiction. So, $L>0$ is impossible. Therefore, $L=0$ is the only possible case. That is, ${lim}_{t\to \mathrm{\infty }}z\left(t\right)=0$. Since $y\left(t\right)$ is bounded, by (C1) we obtain
$\underset{t\to \mathrm{\infty }}{lim}y\left(t\right)=\underset{t\to \mathrm{\infty }}{lim}z\left(t\right)-\underset{t\to \mathrm{\infty }}{lim}P\left(t\right)y\left(t\right)=0$

from (1.2).

Now let us consider the case of $y\left(t\right)<0$ for $t\ge {t}_{1}$. By (1.1) and (1.2),
${z}^{{\mathrm{\Delta }}^{n}}\left(t\right)=-\sum _{i=1}^{m}{Q}_{i}\left(t\right){f}_{i}\left(y\left({\varphi }_{i}\left(t\right)\right)\right)>0$

for $t\ge {t}_{1}$. That is, ${z}^{{\mathrm{\Delta }}^{n}}>0$. It follows that ${z}^{{\mathrm{\Delta }}^{j}}\left(t\right)$ ($j\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$) is strictly monotone and eventually of constant sign. Since $P\left(t\right)$ is an oscillatory function, there exists a ${t}_{2}\ge {t}_{1}$ such that if $t\ge {t}_{2}$, then $z\left(t\right)<0$. Since $y\left(t\right)$ is bounded, by (C1) and (1.2), there is a ${t}_{3}\ge {t}_{2}$ such that $z\left(t\right)$ is also bounded for $t\ge {t}_{3}$. Assume that $x\left(t\right)=-z\left(t\right)$. Then ${x}^{{\mathrm{\Delta }}^{n}}\left(t\right)=-{z}^{{\mathrm{\Delta }}^{n}}\left(t\right)$. Therefore, $x\left(t\right)>0$ and ${x}^{{\mathrm{\Delta }}^{n}}\left(t\right)<0$ for $t\ge {t}_{3}$. Hence, we observe that $x\left(t\right)$ is bounded. Since n is odd, by Lemma 1, there exists a ${t}_{4}\ge {t}_{3}$ and $p=0$ (otherwise $x\left(t\right)$ is not bounded) such that ${\left(-1\right)}^{j}{x}^{{\mathrm{\Delta }}^{j}}\left(t\right)>0$, $j\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$ and $t\ge {t}_{4}$. That is, ${\left(-1\right)}^{j}{z}^{{\mathrm{\Delta }}^{j}}\left(t\right)<0$, $j\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$ and $t\ge {t}_{4}$. In particular, for $t\ge {t}_{4}$ we have ${z}^{\mathrm{\Delta }}\left(t\right)>0$. Therefore, $z\left(t\right)$ is increasing. So, we can assume that ${lim}_{t\to \mathrm{\infty }}z\left(t\right)=L$ ($-\mathrm{\infty }). As in the proof of $y\left(t\right)>0$, we may prove that $L=0$. As for the rest, it is similar to the case of $y\left(t\right)>0$. That is, ${lim}_{t\to \mathrm{\infty }}y\left(t\right)=0$. This contradicts our assumption. Hence the proof is completed. □

Theorem 4 Assume that n is even and (C 1) holds. If the following condition is satisfied:

(C3) There is a function $\phi \left(t\right)$ such that $\phi \left(t\right)\in {C}_{rd}^{1}{\left[{t}_{0},\mathrm{\infty }\right)}_{\mathbb{T}}$. Moreover,
$\underset{t\to \mathrm{\infty }}{lim}sup{\int }_{{t}_{0}}^{t}\phi \left(s\right)\sum _{i=1}^{m}{Q}_{i}\left(s\right)\mathrm{\Delta }s=\mathrm{\infty }$
and
$\underset{t\to \mathrm{\infty }}{lim}sup{\int }_{{t}_{10}}^{t}\frac{{\left[{\phi }^{\mathrm{\Delta }}\left(s\right)\right]}^{2}}{\phi \left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)}\mathrm{\Delta }s<\mathrm{\infty }$

for $\phi \left(t\right)$ and $i=1,2,\dots ,m$. Then every bounded solution of equation (1.1) is oscillatory.

Proof Assume that equation (1.1) has a bounded non-oscillatory solution $y\left(t\right)$. Without loss of generality, assume that $y\left(t\right)$ is eventually positive (the proof is similar when $y\left(t\right)$ is eventually negative). That is, $y\left(t\right)>0$, $y\left(\tau \left(t\right)\right)>0$ and $y\left({\varphi }_{i}\left(t\right)\right)>0$ for $t\ge {t}_{1}\ge {t}_{0}$. By (1.1) and (1.2), we have (3.3) for $t\ge {t}_{1}$. Then ${z}^{{\mathrm{\Delta }}^{n}}\left(t\right)<0$. It follows that ${z}^{{\mathrm{\Delta }}^{j}}\left(t\right)$ ($j\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$) is strictly monotone and eventually of constant sign. Since $P\left(t\right)$ is an oscillatory function, there exists a ${t}_{2}\ge {t}_{1}$ such that for $t\ge {t}_{2}$, we have $z\left(t\right)>0$. Since $y\left(t\right)$ is bounded, by (C1) and (1.2), there is a ${t}_{3}\ge {t}_{2}$, such that $z\left(t\right)$ is also bounded for $t\ge {t}_{3}$. Because n is even, by Lemma 1 when $p=1$ (otherwise $z\left(t\right)$ is not bounded), there exists ${t}_{4}\ge {t}_{3}$ such that for $t\ge {t}_{4}$ we have
${\left(-1\right)}^{j+1}{z}^{{\mathrm{\Delta }}^{j}}\left(t\right)>0,\phantom{\rule{1em}{0ex}}j\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}.$
(3.6)
In particular, since ${z}^{\mathrm{\Delta }}\left(t\right)>0$ for $t\ge {t}_{4}$, $z\left(t\right)$ is increasing. Since $y\left(t\right)$ is bounded,
$\underset{t\to \mathrm{\infty }}{lim}P\left(t\right)y\left(\tau \left(t\right)\right)=0$
by (C1). Let $\delta >1$; i.e., there exists a ${t}_{5}\ge {t}_{4}$ such that by (1.2)
$y\left(t\right)=z\left(t\right)-P\left(t\right)y\left(\tau \left(t\right)\right)>\frac{1}{\delta }z\left(t\right)>0$
for $t\ge {t}_{5}$. We may find a ${t}_{6}\ge {t}_{5}$ such that for $t\ge {t}_{6}$ and $i=1,2,\dots ,m$,
$y\left({\varphi }_{i}\left(t\right)\right)>\frac{1}{\delta }z\left({\varphi }_{i}\left(t\right)\right)>0.$
(3.7)
From (3.3), (3.7) and the properties of f, we have
$\begin{array}{rl}{z}^{{\mathrm{\Delta }}^{n}}\left(t\right)& \le -\sum _{i=1}^{m}{Q}_{i}\left(t\right){f}_{i}\left(\frac{1}{\delta }z\left({\varphi }_{i}\left(t\right)\right)\right)\\ =-\sum _{i=1}^{m}{Q}_{i}\left(t\right)\frac{{f}_{i}\left(\frac{1}{\delta }z\left({\varphi }_{i}\left(t\right)\right)\right)}{z\left({\varphi }_{i}\left(t\right)\right)}z\left({\varphi }_{i}\left(t\right)\right)\end{array}$
(3.8)
for $t\ge {t}_{6}$. Since $z\left(t\right)>0$ is bounded and increasing, ${lim}_{t\to \mathrm{\infty }}z\left(t\right)=L$ ($0). By the continuity of f, we have
$\underset{t\to \mathrm{\infty }}{lim}\frac{{f}_{i}\left(\frac{1}{\delta }z\left({\varphi }_{i}\left(t\right)\right)\right)}{z\left({\varphi }_{i}\left(t\right)\right)}=\frac{{f}_{i}\left(\frac{L}{\delta }\right)}{L}>0.$
Then there is a ${t}_{7}\ge {t}_{6}$ such that for $t\ge {t}_{7}$, $i=1,2,\dots ,m$, we have
$\frac{{f}_{i}\left(\frac{1}{\delta }z\left({\varphi }_{i}\left(t\right)\right)\right)}{z\left({\varphi }_{i}\left(t\right)\right)}\ge \frac{{f}_{i}\left(\frac{L}{\delta }\right)}{2L}=\alpha >0.$
(3.9)
By (3.8), (3.9),
(3.10)
Set
$w\left(t\right)=\frac{{z}^{{\mathrm{\Delta }}^{n-1}}\left(t\right)}{z\left(\frac{1}{\delta }{\varphi }_{i}\left(t\right)\right)}.$
(3.11)
We know from (3.6) that there is a ${t}_{8}\ge {t}_{7}$ such that for a sufficiently large $t\ge {t}_{8}$, $w\left(t\right)>0$. Therefore, Δ-derivating (3.11) we obtain
$\begin{array}{rl}{w}^{\mathrm{\Delta }}\left(t\right)& =\frac{{z}^{{\mathrm{\Delta }}^{n}}\left(t\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}+{z}^{{\mathrm{\Delta }}^{n-1}}\left(\sigma \left(t\right)\right){\left(\frac{1}{z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}\right)}^{\mathrm{\Delta }}\\ =\frac{{z}^{{\mathrm{\Delta }}^{n}}\left(t\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}-\frac{{\delta }^{-1}{\varphi }_{i}^{\mathrm{\Delta }}\left(t\right){z}^{{\mathrm{\Delta }}^{n-1}}\left(\sigma \left(t\right)\right){z}^{\mathrm{\Delta }}\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)z\left({\delta }^{-1}{\varphi }_{i}\left(\sigma \left(t\right)\right)\right)}\end{array}$
(3.12)
$\begin{array}{r}\le \frac{{z}^{{\mathrm{\Delta }}^{n}}\left(t\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}-\frac{{\delta }^{-1}{\varphi }_{i}^{\mathrm{\Delta }}\left(t\right){z}^{{\mathrm{\Delta }}^{n-1}}\left(\sigma \left(t\right)\right){z}^{\mathrm{\Delta }}\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}{{z}^{2}\left({\delta }^{-1}{\varphi }_{i}\left(\sigma \left(t\right)\right)\right)}\\ =\frac{{z}^{{\mathrm{\Delta }}^{n}}\left(t\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}-{\delta }^{-1}{w}^{\sigma }\left(t\right)\frac{{\varphi }_{i}^{\mathrm{\Delta }}\left(t\right){z}^{\mathrm{\Delta }}\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(\sigma \left(t\right)\right)\right)}.\end{array}$
(3.13)
We know from (3.6) that there is a $t\ge {t}_{9}$ such that ${z}^{\mathrm{\Delta }}\left(t\right)>0$ and ${z}^{{\mathrm{\Delta }}^{n-1}}\left(t\right)>0$ for an even n. Since $z\left(t\right)>0$ is increasing $z\left({\delta }^{-1}{\varphi }_{i}\left(\sigma \left(t\right)\right)\right)\ge z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)$ for $i=1,2,\dots ,m$. Therefore, by Lemma 5,
$z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)\ge \lambda {\left(-1\right)}^{n-1}{g}_{n-1}\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right){z}^{{\mathrm{\Delta }}^{n-1}}\left({\varphi }_{i}\left(t\right)\right).$
(3.14)
Then by Δ-derivating (3.14) and using ${g}_{n-1}^{\mathrm{\Delta }}\left(\sigma \left(t\right),t\right)={g}_{n-2}^{\sigma }\left(\sigma \left(t\right),t\right)$, we get
$\begin{array}{rl}{\left[z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)\right]}^{\mathrm{\Delta }}& \ge \lambda {\left(-1\right)}^{n-2}{g}_{n-1}^{\mathrm{\Delta }}\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right){z}^{{\mathrm{\Delta }}^{n-1}}\left({\varphi }_{i}\left(t\right)\right)\\ \ge \lambda {\left(-1\right)}^{n-2}{g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right){z}^{{\mathrm{\Delta }}^{n-1}}\left({\varphi }_{i}\left(t\right)\right)\end{array}$
by Lemma 2
${z}^{\mathrm{\Delta }}\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right){\delta }^{-1}{\varphi }_{i}^{\mathrm{\Delta }}\left(t\right)\ge \lambda {\left(-1\right)}^{n-2}{g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right){z}^{{\mathrm{\Delta }}^{n-1}}\left({\varphi }_{i}\left(t\right)\right).$
Since ${\varphi }_{i}\left(t\right)\le t$, we obtain
${z}^{\mathrm{\Delta }}\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)\ge \frac{\delta \lambda {\left(-1\right)}^{n-2}{g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right){z}^{{\mathrm{\Delta }}^{n-1}}\left(t\right)}{{\varphi }_{i}^{\mathrm{\Delta }}\left(t\right)}.$
(3.15)
Hence by (3.10), (3.14) and (3.15), we have
$\begin{array}{rl}{w}^{\mathrm{\Delta }}\left(t\right)\le & \frac{-\alpha {\sum }_{i=1}^{m}{Q}_{i}\left(t\right)z\left({\varphi }_{i}\left(t\right)\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}\\ -{\delta }^{-1}{w}^{\sigma }\left(t\right)\frac{\delta \lambda {\left(-1\right)}^{n-2}{g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right){z}^{{\mathrm{\Delta }}^{n-1}}\left(t\right)}{{\varphi }_{i}^{\mathrm{\Delta }}\left(t\right)}\frac{{\varphi }_{i}^{\mathrm{\Delta }}\left(t\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(\sigma \left(t\right)\right)\right)}\\ \le & \frac{-\alpha {\sum }_{i=1}^{m}{Q}_{i}\left(t\right)z\left({\varphi }_{i}\left(t\right)\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}\\ -{\delta }^{-1}{w}^{\sigma }\left(t\right)\frac{\delta \lambda {\left(-1\right)}^{n-2}{g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right){\varphi }_{i}^{\mathrm{\Delta }}\left(t\right)}{{\varphi }_{i}^{\mathrm{\Delta }}\left(t\right)}\frac{{z}^{{\mathrm{\Delta }}^{n-1}}\left(\sigma \left(t\right)\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(\sigma \left(t\right)\right)\right)}\\ \le & -\alpha \sum _{i=1}^{m}{Q}_{i}\left(t\right)-\lambda {\left(-1\right)}^{n-2}{g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right){\left({w}^{\sigma }\left(t\right)\right)}^{2},\end{array}$
and then
$\alpha \sum _{i=1}^{m}{Q}_{i}\left(t\right)\le -{w}^{\mathrm{\Delta }}\left(t\right)-\lambda {\left(-1\right)}^{n-2}{w}^{2}\left(t\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right)$
(3.16)
for $t\ge {t}_{10}$. If we multiply (3.16) by $\phi \left(t\right)$ and integrate it from ${t}_{10}$ to t, we obtain by Theorem 2
$\begin{array}{rl}\alpha {\int }_{{t}_{10}}^{t}\phi \left(s\right)\sum _{i=1}^{m}{Q}_{i}\left(s\right)\mathrm{\Delta }s\le & -{\int }_{{t}_{10}}^{t}\phi \left(s\right){w}^{\mathrm{\Delta }}\left(s\right)\mathrm{\Delta }s\\ -{\int }_{{t}_{10}}^{t}\lambda {\left(-1\right)}^{n-2}\phi \left(s\right){w}^{2}\left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)\mathrm{\Delta }s\\ \le & -\left[\phi \left(t\right)w\left(t\right)-\phi \left({t}_{10}\right)w\left({t}_{10}\right)-{\int }_{{t}_{10}}^{t}{\phi }^{\mathrm{\Delta }}\left(s\right){w}^{\sigma }\left(t\right)\mathrm{\Delta }s\right]\\ -{\int }_{{t}_{10}}^{t}\lambda {\left(-1\right)}^{n-2}\phi \left(s\right){w}^{2}\left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)\mathrm{\Delta }s\\ \le & \phi \left({t}_{10}\right)w\left({t}_{10}\right)+{\int }_{{t}_{10}}^{t}{\phi }^{\mathrm{\Delta }}\left(s\right){w}^{\sigma }\left(t\right)\mathrm{\Delta }s\\ -\lambda {\int }_{{t}_{10}}^{t}\phi \left(s\right){w}^{2}\left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)\mathrm{\Delta }s\\ \le & \phi \left({t}_{10}\right)w\left({t}_{10}\right)-\lambda {\int }_{{t}_{10}}^{t}\phi \left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)\\ ×{\left[w\left(s\right)-\frac{{\phi }^{\mathrm{\Delta }}\left(s\right)}{2\lambda \phi \left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)}\right]}^{2}\mathrm{\Delta }s\\ +{\int }_{{t}_{10}}^{t}\frac{{\left[{\phi }^{\mathrm{\Delta }}\left(s\right)\right]}^{2}}{4\lambda \phi \left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)}\mathrm{\Delta }s\\ \le & \phi \left({t}_{10}\right)w\left({t}_{10}\right)+{\int }_{{t}_{10}}^{t}\frac{{\left[{\phi }^{\mathrm{\Delta }}\left(s\right)\right]}^{2}}{4\lambda \phi \left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)}\mathrm{\Delta }s.\end{array}$
Therefore by (C3),
$\begin{array}{rl}\mathrm{\infty }& =\alpha \underset{t\to \mathrm{\infty }}{lim}sup{\int }_{{t}_{10}}^{t}\phi \left(s\right)\sum _{i=1}^{m}{Q}_{i}\left(s\right)\mathrm{\Delta }s\\ \le \phi \left({t}_{10}\right)w\left({t}_{10}\right)+\frac{1}{4\lambda }\underset{t\to \mathrm{\infty }}{lim}sup{\int }_{{t}_{10}}^{t}\frac{{\left[{\phi }^{\mathrm{\Delta }}\left(s\right)\right]}^{2}}{\phi \left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)}\mathrm{\Delta }s\\ <\mathrm{\infty }.\end{array}$

This is a contradiction.

Now let us consider the case of $y\left(t\right)<0$ for $t\ge {t}_{1}$. By (1.1) and (1.2), we have
${z}^{{\mathrm{\Delta }}^{n}}\left(t\right)=-\sum _{i=1}^{m}{Q}_{i}\left(t\right){f}_{i}\left(y\left({\varphi }_{i}\left(t\right)\right)\right)>0$

for $t\ge {t}_{1}$. That is, ${z}^{{\mathrm{\Delta }}^{n}}>0$. It follows that ${z}^{{\mathrm{\Delta }}^{j}}\left(t\right)$ ($j\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$) is strictly monotone and eventually of constant sign. Since $P\left(t\right)$ is an oscillatory function, there exists a ${t}_{2}\ge {t}_{1}$ such that $z\left(t\right)<0$ for $t\ge {t}_{2}$. Since $y\left(t\right)$ is bounded, by (C1) and (1.2), there is a ${t}_{3}\ge {t}_{2}$ such that $z\left(t\right)$ is also bounded for $t\ge {t}_{3}$. Assume that $x\left(t\right)=-z\left(t\right)$. Then ${x}^{{\mathrm{\Delta }}^{n}}\left(t\right)=-{z}^{{\mathrm{\Delta }}^{n}}\left(t\right)$. Therefore, $x\left(t\right)>0$ and ${x}^{{\mathrm{\Delta }}^{n}}\left(t\right)<0$ for $t\ge {t}_{3}$. Hence, we observe that $x\left(t\right)$ is bounded. Since n is odd, by Lemma 1, there exists a ${t}_{4}\ge {t}_{3}$ and $p=1$ (otherwise $x\left(t\right)$ is not bounded) such that ${\left(-1\right)}^{k}{x}^{{\mathrm{\Delta }}^{k}}\left(t\right)>0$, $k\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$ and $t\ge {t}_{4}$. That is, ${\left(-1\right)}^{k}{z}^{{\mathrm{\Delta }}^{k}}\left(t\right)<0$, $k\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$ and $t\ge {t}_{4}$. In particular, for $t\ge {t}_{4}$ we have ${z}^{\mathrm{\Delta }}\left(t\right)>0$. Therefore, $z\left(t\right)$ is increasing. For the rest of the proof, we can proceed the proof similarly to the case of $y\left(t\right)>0$. Hence, the proof is completed. □

## Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Sciences and Arts, Usak University, 1 Eylul Campus, Usak, 64200, Turkey
(2)
Department of Mathematics, Faculty of Sciences and Literatures, Kastamonu University, Kastamonu, Turkey

## References 